Simplify; express your answer in exponential form. Assume $q\neq 0, t\neq 0$. $\dfrac{{(q^{-4})^{-4}}}{{(q^{2}t^{3})^{-1}}}$
Explanation: To start, try working on the numerator and the denominator independently. In the numerator, we have ${q^{-4}}$ to the exponent ${-4}$ . Now ${-4 \times -4 = 16}$ , so ${(q^{-4})^{-4} = q^{16}}$ In the denominator, we can use the distributive property of exponents. ${(q^{2}t^{3})^{-1} = (q^{2})^{-1}(t^{3})^{-1}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(q^{-4})^{-4}}}{{(q^{2}t^{3})^{-1}}} = \dfrac{{q^{16}}}{{q^{-2}t^{-3}}}$ Break up the equation by variable and simplify. $\dfrac{{q^{16}}}{{q^{-2}t^{-3}}} = \dfrac{{q^{16}}}{{q^{-2}}} \cdot \dfrac{{1}}{{t^{-3}}} = q^{{16} - {(-2)}} \cdot t^{- {(-3)}} = q^{18}t^{3}$.